 # Compute the standard deviation for the pretest and posttest scores that were used in problems 2.6 and 3.12. The scores are reproduced here. Taking into account all of the information you have on these variables, write a paragraph describing how the sample changed from test to test. What does the standard deviation add to the information you already had? Case Pretest Posttest A 8 12 B 7 13 C 10 12 D 15 19 E 10 8 F 10 17 G 3 12 H 10 11 I 5 7 J 15 12 K 13 20 L 4 5 M 10 15 N 8 11 O 12 20 2.6 S W A local youth service agency has begun a sex education program for teenage girls who have been referred by the juvenile courts. The girls were given a 20-item test for general knowledge about sex, contraception, and anatomy and physiology upon admission to the program and again after completion of the program. The scores of the first 15 girls to complete the program are as follows. Case Pretest Posttest Case Pretest Posttest A 8 12 I 5 7 B 7 13 J 15 12 C 10 12 K 13 20 D 15 19 L 4 5 E 10 8 M 10 15 F 10 17 N 8 11 G 3 12 O 12 20 H 10 11 3.12 S W Compute the median and mean for both the pretest and posttest for the test scores first presented in problem 2.6 and reproduced here. Interpret the statistics. Case Pretest Posttest A 8 12 B 7 13 C 10 12 D 15 19 E 10 8 F 10 17 G 3 12 H 10 11 I 5 7 J 15 12 K 13 20 L 4 5 M 10 15 N 8 11 O 12 20 ### Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836 ### Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836

#### Solutions

Chapter
Section
Chapter 4, Problem 4.6P
Textbook Problem
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## Compute the standard deviation for the pretest and posttest scores that were used in problems 2.6 and 3.12. The scores are reproduced here. Taking into account all of the information you have on these variables, write a paragraph describing how the sample changed from test to test. What does the standard deviation add to the information you already had? Case Pretest Posttest A 8 12 B 7 13 C 10 12 D 15 19 E 10 8 F 10 17 G 3 12 H 10 11 I 5 7 J 15 12 K 13 20 L 4 5 M 10 15 N 8 11 O 12 20 2.6 S W A local youth service agency has begun a sex education program for teenage girls who have been referred by the juvenile courts. The girls were given a 20-item test for general knowledge about sex, contraception, and anatomy and physiology upon admission to the program and again after completion of the program. The scores of the first 15 girls to complete the program are as follows. Case Pretest Posttest Case Pretest Posttest A 8 12 I 5 7 B 7 13 J 15 12 C 10 12 K 13 20 D 15 19 L 4 5 E 10 8 M 10 15 F 10 17 N 8 11 G 3 12 O 12 20 H 10 11 3.12 S W Compute the median and mean for both the pretest and posttest for the test scores first presented in problem 2.6 and reproduced here. Interpret the statistics. Case Pretest Posttest A 8 12 B 7 13 C 10 12 D 15 19 E 10 8 F 10 17 G 3 12 H 10 11 I 5 7 J 15 12 K 13 20 L 4 5 M 10 15 N 8 11 O 12 20

Expert Solution
To determine

To find:

The standard deviation for the pretest and posttest scores

### Explanation of Solution

Given:

The following table shows the data of pretest and posttest scores of the teenage girls.

 Case Pretest Posttest A 8 12 B 7 13 C 10 12 D 15 19 E 10 8 F 10 17 G 3 12 H 10 11 I 5 7 J 15 12 K 13 20 L 4 5 M 10 15 N 8 11 O 12 20

Formula used:

Let the data values be Xi’s.

The formula to calculate mean is given by,

X¯=i=1NXiN

The formula to calculate standard deviation is given by,

s=(XiX¯)2N

Where, N is the population size.

Calculation:

For pretest scores:

The size of the population is 15.

The mean is given by,

X¯=i=1NXiN

Substitute 15 for N, 8 for X1, 7 for X2 and so on in the above mentioned formula,

X¯=8+7+............+8+1215=14015=9.33333 ……(1)

Consider the following table of sum of squares,

 Scores (Xi) (Xi−X¯) (Xi−X¯)2 8 −1.3333 1.777769 7 −2.3333 5.444429 10 0.66667 0.444449 15 5.66667 32.11115 10 0.66667 0.444449 10 0.66667 0.444449 3 −6.3333 40.11107 10 0.66667 0.444449 5 −4.3333 18.77775 15 5.66667 32.11115 13 3.66667 13.44447 4 −5.3333 28.44441 10 0.66667 0.444449 8 −1.3333 1.777769 12 2.66667 7.111129 ∑(Xi−X¯)=0 ∑(Xi−X¯)2=183.333

From equation (1), substitute 8 for X1 and 9.33333 for X¯ in (X1X¯).

(X1X¯)=(89.33333)(X1X¯)=1.33333

Square the both sides of the equation.

(X1X¯)2=(1.3333)2(X1X¯)2=1.777769

Proceed in the same manner to calculate (XiX¯)2 for all the 1iN for the rest data and refer table for the rest of the (XiX¯)2 values calculated. Then the value of (XiX¯)2 is calculated as,

(XiX¯)2=1.777769+5.444429+............+7.111129=183.333 ……(2)

The standard deviation is given by,

s=(XiX¯)2N

From equation (2), substitute 183.333 for (XiX¯)2 and 15 for N in the above mentioned formula,

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